X
bounds another minimal disc which lies close to S . Form a
Jordan curve P ^ by connecting P, and with two arcs nearly
parallel to a longitude. Let be the flat disc bounded by jfi^.
Then 1"^ bounds a stable minimal disc consisting of discs close
to D, and joined by a bridge, and 1"^ also bounds a stable
minimal disc Dj. consisting of discs close to D 1 and joined by
a bridge. Let P^ be a Jordan curve close to but not intersecting
P x. Then 1“^ and bound stable minimal discs that intersect,
and these may be connected by a bridge. As with the example for
Question 1, we m a y prove that such a discAexists by constructing
a manifold satisfying condition
(c),
which in this case is immersed in (R* rather than embedded. (The reason for the discSt.
Remarks on Question 2. We are not using the full strength of Theorem 4,
since we do n o t need A to be embedded.
Having constructed an immersed minimal disc A with c S a
i t is natural to ask whether such a disc can have a branch point.
I t does not seem possible to construct a disc with a branch point
u s ing the bridge principle.
The Jordan curve "bA appears to bound at least five stable
minimal discs, four obtained by the bridge principle together with
the disc of least area, which appears to be a "ribbon" close to the
curve C; it appears that only one of these fails to be embedded.
There should be at least four unstable minimal discs as well. Given
that a Jordan curve
P
bounds a minimal disc that is not embedded, h o w many other minimal discs mustP
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